the sum of first n term …

Here Sum of the n terms of an Arithmetic progression, Sn = 5n² + 3n
S₁ = 5 {tex}\times{/tex} 1² + 3 {tex}\times{/tex} 1 = 8 = t₁ ....(i)
S₂ = 5 {tex}\times{/tex} 2² + 3 {tex}\times{/tex} 2 = 26 = t₁ + t₂ ...(ii)
S₃ = 5 {tex}\times{/tex} 3² + 3 {tex}\times{/tex} 3 = 54 = t₁ + t₂ + t₃ ...(iii)
From (i), (ii) and (iii),
t₁ = 8, t₂ = 18, t₃ = 28
So Common difference, d = 18 - 8 = 10 and first term a=8.
Now t_m = 168 (given)

We know that {tex}{T_m}=a+(m-1)d{/tex}
{tex}\Rightarrow{/tex} a + (m - 1)d = 168
{tex}\Rightarrow{/tex} 8 + (m - 1) {tex}\times{/tex} 10 = 168
{tex}\Rightarrow{/tex} (m - 1) {tex}\times{/tex} 10 = 160

{tex}\implies{/tex} m-1 ={tex}\frac{160}{10}{/tex}
{tex}\Rightarrow{/tex} m - 1 = 16
Therefore,  m = 17
So, t₂₀ = a  + (20 - 1)d
= 8 + 19 {tex}\times{/tex} 10
= 8 + 190
= 198